Answer:
<em>O</em><em>r</em><em>g</em><em>a</em><em>n</em><em>i</em><em>s</em><em>m</em><em> </em><em>,</em><em> </em><em>P</em><em>o</em><em>p</em><em>u</em><em>l</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>,</em><em> </em><em>C</em><em>o</em><em>m</em><em>m</em><em>u</em><em>n</em><em>i</em><em>t</em><em>y</em><em> </em><em>,</em><em> </em><em>E</em><em>c</em><em>o</em><em>s</em><em>y</em><em>s</em><em>t</em><em>e</em><em>m</em><em> </em><em>.</em>
Explanation:
<u>Organism</u> is one type of species .
<u>Population</u> is a group of same species living together .
<u>Community</u> is a group of different species living together .
<u>Ecosystem</u> is a large group of community of species and its environment consisting of biotic and abiotic factor .
Answer:
He blasted alpha rays at the gold foils. The Alpha rays are charged positive. When he blasted them the protons deflected and changed the course of them and the electrons surrounding the atom let them through. In the plum pudding model they were supposed to be sprinkled on which was clearly wrong according to that experiment.
Explanation:
Answer:
B. To change from a liquid state to a solid state is called Freezing
Answer:
The molarity of the solution: 0,27M
Explanation:
First , we calculate the weight of 1 mol of NaCl:
Weight 1mol NaCl= Weight Na + Weight Cl= 23 g+ 35, 5 g= 58, 5 g/mol
58,5 g---1 mol NaCl
64 g--------x= (64 g x1 mol NaCl)/58,5 g= 1, 09 mol NaCl
A solution molar--> moles of solute in 1 L of solution:
4 L-----1,09 mol NaCl
1L----x0( 1L x1,09 mol NaCl)/4L =0,27moles NaCl--->0,27M
Answer:
The activation energy for this reaction = 23 kJ/mol.
Explanation:
Using the expression,
Where,
is the activation energy
R is Gas constant having value = 8.314×10⁻³ kJ / K mol
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (280 + 273.15) K = 553.15 K
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (376 + 273.15) K = 649.15 K
So,
<u>The activation energy for this reaction = 23 kJ/mol.</u>
